Empirical and Molecular Formulae
1
Find the empirical formulae of the substances with the following percentage
composition
(a) 80% copper and 20% oxygen
(b) 53% aluminium and 47% oxygen
(c) 1·6% hydrogen, 22·2% nitrogen and 76·2% oxygen
2
Calculate the empirical formula of the oxide of sulfur which is 40% sulfur by mass.
3
A hydrocarbon contains 90% carbon. Calculate the empirical formula of the
hydrocarbon.
4
An oxide of silicon was produced from 0·28 g of silicon. The mass of the oxide
was 0·60 g. Calculate the empirical formula of the oxide.
5
10·2 g of vanadium is combined with 21·3 g of chlorine to make vanadium
chloride. Calculate the empirical formula of the vanadium chloride.
6
The empirical formula of a substance is CH2. Its molar mass is 84 g mol−1. Find
the molecular formula of the substance.
7
Analysis gave the following data for three unknown substances
(a) 85·7% C and 14·3% H; M = 28 g mol−1
(b) 30·4% N and 69·6% O; M = 92 g mol−1
(c) 2% H, 33% S and 65% O; M = 98 g mol−1
Find the molecular formulae for each of the substances
8
A hydrocarbon contains 82·7% carbon and 17·3% hydrogen by mass
(a) What is the empirical formula of the hydrocarbon?
(b) Given that the molar mass of the hydrocarbon is 58 g mol−1, what is its
molecular formula?
9
A gaseous hydrocarbon was found to contain 80% carbon and 20% hydrogen by
mass
(a) Calculate the empirical formula of the hydrocarbon
(b) The hydrocarbon was found to have a molar mass of 30 g mol−1. Use this
value to work out the molecular formula of the hydrocarbon.
10
(a) An oxide of nitrogen contains 30·3% nitrogen. Calculate the empirical
formula of the oxide.
(b) If the molar mass of the oxide is 92 g mol−1, use your answer in (a) to
determine the molecular formula of the oxide.
11
(a) Vitamin C has a mass composition of 40·92% carbon, 4·58% hydrogen and
54·40% oxygen. Calculate the empirical formula of Vitamin C.
(b) If the molar mass of Vitamin C is 176 g mol−1, use your answer in (a) to
determine the molecular formula for Vitamin C.
12
(a) An organic compound has a mass composition if 26·67% carbon, 2·22%
hydrogen and 71·11% oxygen. Calculate the empirical formula of the
compound.
(b) If the molar mass of the compound is 90 g mol−1, use your answer in (a) to
determine the molecular formula of the compound.
13
A lead sulfide compound was made by heating 2·95 g of lead with excess sulfur.
Once all the lead had reacted, the excess sulfur was burnt off. The mass of the
final product was 3·42 g.
M(Pb) = 207 g mol−1, M(S) = 32·1 g mol−1
(a) Calculate the moles of lead used.
(b) Calculate to mass and hence the moles of sulfur reacted.
(c) Calculate the empirical formula of the compound.
Activity 7E: Percentage Composition, Empirical and Molecular
Formulae Answers
1
copper
%
80
M
63·5
n
1·26
÷ smallest
1·25
simplest ratio
1
empirical formula: CuO
(a)
2
3
oxygen
20
16·0
1·25
1·25
1
:
(b)
aluminium
%
53
M
27·0
n
1·96
÷ smallest
1·96
×2
1
:
simplest ratio
2
empirical formula: Al2O3
(c)
hydrogen
%
1·6
M
1·0
n
1·6
÷ smallest
1·59
:
simplest ratio
1
empirical formula: HNO3
sulfur
%
40
M
32·1
n
1·25
÷ smallest
1·25
simplest ratio
1
empirical formula: SO3
carbon
%
90
M
12·0
n
7·5
÷ smallest
7·5
×3
1
simplest ratio
3
empirical formula: C3H4
oxygen
47
16·0
2·94
1·96
1·50
3
nitrogen
22·2
14·0
1·59
1·59
1
:
oxygen
60
16·0
3·75
1·25
3
:
hydrogen
10
1·0
10
7·5
1·33 (1⅓)
4
:
oxygen
76·2
16·0
4·76
1·59
3
4
silicon
oxygen
m
0·28
0·60 − 0·28
M
28·1
16·0
−3
n
9·96 × 10
0·0200
−3
÷ smallest
9·96 × 10
9·96 × 10−3
:
simplest ratio
1
2
empirical formula: SiO2
5
vanadium
m
10·2
M
50·9
n
0·200
÷ smallest
0·200
simplest ratio
1
empirical formula: VCl3
6
:
chlorine
21·3
35·5
0·600
0·200
3
M(molecule )
84
 empiricalf ormula =
 CH2 = 6CH2
M(empirical )
14  0
= C6H12
molecular formula =
7
(a)
carbon
hydrogen
%
85·7
14·3
M
12·0
1·0
n
7·14
14·3
÷ smallest
7·14
7·14
:
simplest ratio
1
2
empirical formula: CH2
28
molecular formula =
CH 2 = 2 × CH2 = C2H4
14
7
(b)
nitrogen
oxygen
%
30·4
69·6
M
14·0
16·0
n
2·17
4·35
÷ smallest
2·17
2·17
:
simplest ratio
1
2
empirical formula: NO2
92
molecular formula =
NO 2 = 2 × NO2 = N2O4
46  0
(c)
hydrogen
2
1·0
2
1·03
2
%
M
n
÷ smallest
simplest ratio
:
sulfur
33
32·1
1·03
1·03
1
:
oxygen
65
16·0
4·06
1·03
4
empirical formula: H2SO4
98
molecular formula =
H 2SO4 = 1 × H2SO4 = H2SO4
98  1
8
(a)
carbon
%
82·7
M
12·0
n
6·89
÷ smallest
6·89
×2
1
simplest ratio
2
empirical formula: C2H5
(b) molecular formula =
9
(a)
58
C 2 H 5 = 2 × C2H5 = C4H10
29  0
carbon
%
80
M
12·0
n
6·67
÷ smallest
6·67
simplest ratio
1
empirical formula: CH3
(b) molecular formula =
hydrogen
17·3
1·0
17·3
6·89
2·51
:
5
:
hydrogen
20
1·0
20
6·67
3
30
CH3 = 2 × CH3 = C2H6
15  0
10
(a)
nitrogen
%
30·3
M
14·0
n
2·16
÷ smallest
2·16
simplest ratio
1
empirical formula: NO2
(b) molecular formula =
11
(a)
12
(a)
13
(a) n(Pb) =
:
hydrogen
4·58
1·0
4·58
3·41
1·34 (1⅓)
4
:
oxygen
54·50
16·0
3·41
3·41
1
3
176
C3H 4 O3 = 2 × C3H4O3 = C6H8O6
88  0
carbon
%
26·67
M
12·0
n
2·22
÷ smallest
2·22
simplest ratio
1
empirical formula: CHO2
(b) molecular formula =
:
92
NO 2 = 2 × NO2 = N2O4
46  0
carbon
%
40·92
M
12·0
n
3·41
÷ smallest
3·41
×3
1
simplest ratio
3
empirical formula: C3H4O3
(b) molecular formula =
oxygen
69·7
16·0
4·36
2·16
2
hydrogen
2·22
1·0
2·22
2·22
:
1
:
oxygen
71·11
16·0
4·44
2·22
2
90
CHO 2 = 2 × CHO2 = C2H2O4
45  0
m(Pb)
2  95
=
= 0·01425 mol
M(Pb)
207
(b) m(S) = m(lead sulfide) – m(lead) = 3·42 − 2·95 = 0·47 g
n(S) =
(c)
m(S) 0  47
=
= 0·0146 mol
M (S)
32  1
n (S)
0  0146
1
=
= , i.e. empirical formula is PbS
n (Pb) 0  01425 1